CorollaryStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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A nonintegral order bounds the canonical genus by its floor

Statement

Let $f$ be a nonzero entire function of finite nonintegral order $ρ$ , and let $(a_{n})_{n \geq 1}$ be its nonzero zero sequence. Then the canonical genus of $(a_{n})$ is at most $⌊ ρ ⌋$ .

Facts & Assumptions

Given: A nonzero entire function $f$ of finite nonintegral order $ρ$ and its nonzero zero sequence $(a_{n})$ .

[F1]

Hadamard factorization writes $f$ as an exponential of a polynomial times the genus- $⌊ ρ ⌋$ canonical product over its nonzero zeros (Hadamard factorization for finite-order entire functions).

Proof

technique · direct

1.1

By [F1], the genus- $⌊ ρ ⌋$ canonical product $\prod_{n \geq 1} E_{⌊ ρ ⌋} (z / a_{n})$ already converges.

F1given

2.1

By definition, the canonical genus is the least integer for which the corresponding canonical product converges. Step 1.1 therefore gives $canonical genus \leq ⌊ ρ ⌋$ . The nonintegrality of $ρ$ makes $⌊ ρ ⌋$ the largest integer strictly below $ρ$ , which is the usual Hadamard bound.

F1step 1.1algebra∎

Depends on

Hadamard factorization for finite-order entire functions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matthias Weber, Complex Analysis, Ch. 3 §3.6 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Hadamard's factorization theorem (standard reference, not scraped)